BZOJ 4756: [Usaco2017 Jan] Promotion Counting segment tree merge

title

BZOJ 4756

LUOGU 3605

Simplify the meaning of the questions:

\ (n-\) cows company constitute a tree, each cow has a capacity value \ (P_i \) , \ (. 1 \) Number of cows root.
Q. For each cow, its sub-tree has several value is greater than its capacity.

analysis

Segment tree merger bare title, so learning about, be released devil.

Reference: Mychael .

code

#include<bits/stdc++.h>

const int maxn=1e5+10;

namespace IO
{
    char buf[1<<15],*fs,*ft;
    inline char getc() { return (ft==fs&&(ft=(fs=buf)+fread(buf,1,1<<15,stdin),ft==fs))?0:*fs++; }
    template<typename T>inline void read(T &x)
    {
        x=0;
        T f=1, ch=getchar();
        while (!isdigit(ch) && ch^'-') ch=getchar();
        if (ch=='-') f=-1, ch=getchar();
        while (isdigit(ch)) x=(x<<1)+(x<<3)+(ch^48), ch=getchar();
        x*=f;
    }

    char Out[1<<24],*fe=Out;
    inline void flush() { fwrite(Out,1,fe-Out,stdout); fe=Out; }
    template<typename T>inline void write(T x,char str)
    {
        if (!x) *fe++=48;
        if (x<0) *fe++='-', x=-x;
        T num=0, ch[20];
        while (x) ch[++num]=x%10+48, x/=10;
        while (num) *fe++=ch[num--];
        *fe++=str;
    }
}

using IO::read;
using IO::write;

namespace SGT
{
    struct Orz{int l,r,z;}c[maxn*30];
    int cnt=0;
    inline void Change(int &x,int l,int r,int k)
    {
        if (!x) x=++cnt;
        ++c[x].z;
        if (l==r) return ;
        int mid=(l+r)>>1;
        if (k<=mid) Change(c[x].l,l,mid,k);
        else Change(c[x].r,mid+1,r,k);
        c[x].z=c[c[x].l].z+c[c[x].r].z;
    }

    inline int query(int x,int l,int r,int tl,int tr)
    {
        if (!x || tr<tl) return 0;
        if (tl<=l && r<=tr) return c[x].z;
        int mid=(l+r)>>1, ans=0;
        if (tl<=mid) ans+=query(c[x].l,l,mid,tl,tr);
        if (tr>mid) ans+=query(c[x].r,mid+1,r,tl,tr);
        return ans;
    }

    inline int merge(int x,int y)
    {
        if (!x) return y;
        if (!y) return x;
        c[x].l=merge(c[x].l,c[y].l);
        c[x].r=merge(c[x].r,c[y].r);
        c[x].z=c[c[x].l].z+c[c[x].r].z;
        return x;
    }
}

using SGT::Change;
using SGT::query;
using SGT::merge;

int ver[maxn],Next[maxn],head[maxn],len;
inline void add(int x,int y)
{
    ver[++len]=y,Next[len]=head[x],head[x]=len;
}

int tot,n;
int ans[maxn],val[maxn],rt[maxn];
inline void dfs(int x)
{
    for (int i=head[x]; i; i=Next[i])
    {
        int y=ver[i];
        dfs(y);
        rt[x]=merge(rt[x],rt[y]);
    }
    ans[x]=query(rt[x],1,n,val[x]+1,tot);
}

int a[maxn];
int main()
{
    read(n);
    for (int i=1; i<=n; ++i) read(val[i]),a[i]=val[i];
    for (int i=2,x; i<=n; ++i) read(x),add(x,i);
    std::sort(a+1,a+n+1);
    tot=std::unique(a+1,a+n+1)-a-1;
    for (int i=1; i<=n; ++i) val[i]=std::lower_bound(a+1,a+tot+1,val[i])-a;
    for (int i=1; i<=n; ++i) Change(rt[i],1,n,val[i]);
    dfs(1);
    for (int i=1; i<=n; ++i) write(ans[i],'\n');
    IO::flush();
    return 0;
}